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Natali [406]
3 years ago
14

If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz

ontal, how much work does the child do on the wagon?
Physics
1 answer:
lilavasa [31]3 years ago
4 0

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

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A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0
3 years ago
A ball that is attached to a string travels in a horizontal, circular path, as shown in Figure 1. At time t0 , the ball has a sp
Levart [38]

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

<u>F₁ = 4 F₀</u>

5 0
3 years ago
A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

7 0
3 years ago
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Which of the following is an example of a physical change?
Stels [109]

Answer:

1 question is- I believe 4. Melting Question 2 is Density

Explanation:

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7 0
4 years ago
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Furkat [3]

Answer:

6

Explanation:

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7 0
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